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$\newcommand{\bf}[1]{\mathbb #1}\newcommand{\sc}[1]{\mathscr #1}$ A duality between two vector spaces $E$ and $F$ over $\bf K$ ($= {\bf R}$ of ${\bf C}$) is, by definition, a bilinear form $$ \langle \cdot, \cdot\rangle :E\times F\to \bf K, $$ such that, if $\langle x, y\rangle =0$ for every $x$ in $E$, then $y=0$. And vice-versa.

Given a duality as above, one defines the weak topology on $F$, usually denoted $\sigma (F,E)$, to be the coarsest topology according to which the linear functionals $$ y\in F\mapsto \langle x, y\rangle \in \bf K $$ are continuous for every $x$ in $E$.

It is a classical fact that every $\sigma (F,E)$-continuous linear functional $\varphi :F\to \bf K$, may be represented by a vector in $E$ in the sense that there exists a (necessarily unique) $x$ in $E$ such that $$ \phi(y) = \langle x, y\rangle ,\quad\forall y\in E. $$

One could therefore ask:

Question. Does the above still hold if continuity is replaced by sequential continuity. In other words, must every sequentially $\sigma (F, E)$-continuous linear functional on $F$ be represented by a vector in $E$.

Before the reader jumps to the task of proving or disproving it, let me say that unfortunately the answer is negative, a counter example being presented below.

So let me specialize this a bit by restricting to the situation in which $E$ is a Banach space and $F$ is its topological dual, with the canonical duality $$ \langle x, \varphi \rangle = \varphi (x), \quad \forall x\in E, \quad \forall \varphi \in E'. $$

To be precise:

Question. Let $E$ be a Banach space and let $\varphi $ be a linear functional on $E'$ which is sequentially $\sigma (E',E)$-continuous. Is $\varphi $ necessarily represented by a vector in $E$?

This is obviously true if $E$ is reflexive and I think I can also prove it for $E=c_0$, as well as for $E=\ell ^1$.


A COUNTER EXAMPLE

Let $E=\sc F(H)$ be the set of all finite-rank operators on Hilbert's space, and $F=\sc B(H)$, with duality defined by means of the trace, namely $$ \langle S, T\rangle = \text{tr}(ST), \quad\forall S\in \sc F(H), \quad\forall T\in \sc B(H). $$

In this case $\sigma \big (\sc B(H),\sc F(H)\big )$ turns out to be the weak operator topology (WOT), which coincides with the sigma weak operator topology ($\sigma $-WOT) on bounded subsets of $\sc B(H)$.

Since WOT-convergent sequences are bounded by Banach-Steinhauss, we have that the WOT-convergent sequences are the same as the $\sigma $-WOT convergent ones. It follows that every $\sigma $-WOT-continuous linear functional on $\sc B(H)$ is also WOT-continuous. Making a long story short, for every trace class operator $S$ on $H$ of infinite rank, the linear functional $$ T\in \sc B(H) \mapsto \text{tr}(ST)\in {\bf C} $$ is sequentially WOT-continuous, but it is not represented by an operator in $\sc F(H)$.

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    $\begingroup$ This is at least true if $E$ is separable. Indeed, in that case, by the Krein-Smulian theorem, a convex subset of $E'$ is $\sigma(E',E)$-closed if and only if it is $\sigma(E',E)$ sequentially closed. See Corollary V.12.7 in Conway's course in functional analysis. Apply this to the kernel of $\varphi$. $\endgroup$ – Mikael de la Salle Jan 10 at 19:42
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    $\begingroup$ @Mikael, thanks for pointing it out. In fact the following Corollary in Conway's book, namely V.12.8, is exactly what I am asking! $\endgroup$ – Ruy Jan 10 at 19:58
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    $\begingroup$ Indeed, I had missed this Corollary. Note however that separability of $E$ is assumed, and used essentially in the proof. $\endgroup$ – Mikael de la Salle Jan 10 at 20:03
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    $\begingroup$ Related: mathoverflow.net/q/284276/61785 $\endgroup$ – Robert Furber Jan 11 at 7:57
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    $\begingroup$ Side note: These questions can bump into set-theoretic undecidability. Consider $\ell^\infty(\mathbb{R})$ (the bounded, not required to be measurable, functions $\mathbb{R} \rightarrow \mathbb{R}$) as the dual space of $\ell^1(\mathbb{R})$. There is a sequentially weak-* continuous functional that is not weak-* continuous (i.e. a point evaluation) if and only if there is an atomless real-valued measurable cardinal (i.e. $(\mathbb{R}, \mathcal{P}(\mathbb{R}))$ admits a probability measure vanishing on singletons). $\endgroup$ – Robert Furber Jan 11 at 8:36
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Mikael de la Salle points out this is true when $E$ is separable, as shown in Corollary V.12.8 of Conway, A Course in Functional Analysis, 2e.

For a non-separable counterexample, consider the uncountable ordinal space $[0, \omega_1]$, which is compact Hausdorff, and $E = C([0, \omega_1])$. By the Riesz representation theorem, $E'$ is the space of signed Radon measures $\mu$ on $[0, \omega_1]$ with its total variation norm. Let $\varphi(\mu) = \mu(\{\omega_1\})$. This is clearly not represented by any vector in $E$ since the function $1_{\{\omega_1\}}$ is not continuous, but I claim $\varphi$ is sequentially $\sigma(E', E)$ continuous.

Let $\mu_n$ be a sequence converging to 0 in $\sigma(E', E)$ and fix $\epsilon > 0$. Since each $\mu_n$ is Radon, so is its total variation measure $|\mu_n|$, and thus we can approximate $\{\omega_1\}$ in $|\mu_n|$-measure from outside by open sets. So there exists $\alpha_n < \omega_1$ such that $|\mu_n|((\alpha_n, \omega_1)) < \epsilon$. Let $\alpha = \sup_n \alpha_n < \omega_1$; then $|\mu_n((\alpha, \omega_1))| \le |\mu_n|((\alpha, \omega_1)) < \epsilon$ for every $n$.

Define $f : [0, \omega_1] \to \mathbb{R}$ by $$f(x) = \begin{cases} 0, & x \le \alpha \\ 1, & x > \alpha \end{cases}$$ and note that $f$ is continuous. Now $$\varphi(\mu_n) = \mu_n(\{\omega_1\}) = \mu_n((\alpha, \omega_1]) - \mu_n((\alpha, \omega_1)) = \int f\,d\mu_n - \mu_n((\alpha, \omega_1)).$$

But by assumption $\int f\,d\mu_n \to 0$, and $|\mu_n((\alpha, \omega_1))| < \epsilon$, so we conclude $\varphi(\mu_n) \to 0$.

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